# A question on the lorentz transforms and special relativity

## Homework Statement

Not sure whether to put this in advanced or intro forum, but ill just put it in here.

So heres the context of the question:
a linear swarm of 3 meteors are travelling at a speed of 0.9c towards our solar system (in the solar system reference frame). Each adjacent meteor is separated by a distance of 0.5 light hours in the swarm reference frame and this distance is maintained throughout their journey. These meteors pass the first beacon of a grid of solar system beacons, these beacons each have their own clocks and are synchronized with eachother, the frame of the solar system beacons system will be denoted SSBS. Each meteor also has an 'internal clock' and as the first meteor passes the first beacon, it synchronizes its clock with the beacons clock and somehow orchestrates the other meteors to synchronize their clocks with the first meteor and this event (when the first meteor passes the first beacon) is deemed the origin in both the swarm frame and the SSBS frame. After this the meteors continue their journey which is headed for Earth, a distance of 10 light hours away in the SSBS ref frame. (Disregard any general relativity effects for this question, all frames are inertial frames).

And here is the question im having a little trouble with:

(d) At the instant that the first meteor in the swarm arrives at Earth, what are the space time co ordinates (x,t) in the SSBS frame of all the meteors in the swarm? give your co ordinates in light hours and hours.

## Homework Equations

Inverse Lorentz transformations, length contraction formula and time dilation formula.

## The Attempt at a Solution

So first I calculated the contracted distance between the first beacon and Earth according to the swarm of meteors, and this came out to be 4.3589 light hours. From this i calculated the proper time (the time in the swarm ref frame) between the two events (passing the first beacon and arriving at Earth) this came out to be 4.843 hours. Then I found the time dilation in the SSBS frame for the two events, and it was 11.11 hours.

Next thing i did was find the co ordinates of all the meteors in the swarm frame when the first meteor arrived at Earth. These were their co ordinates in the form (x,t) in light hours and hours;
Meteor 1: (0, 4.843)
meteor 2: (-0.5, 4.843)
meteor 3: (-1, 4.843)
Since the spatial origin is always at the first meteor in the swarm frame.

Now all i need to do is use the inverse lorentz transformations, right?

So; using ## x=\gamma(x'+vt') ## and ##t=\gamma(t'+\frac{vx'}{c^2}) ##
##\gamma_{0.9} = 2.294##

Meteor 1: x = 2.294(0 + 0.9*4.843) = 10 Light hours
t = 2.294(4.843 + 0) = 11.11 hours since x'=0

Meteor 2: x = 2.294(-0.5 + 0.9*4.843) = 8.852 Light hours
t = 2.294(4.843 - 0.9*0.5) = 10.08 hours

Meteor 3: x = 2.294(-1 + 0.9*4.843) = 7.705 Light hours
t = 2.294(4.843 - 0.9*1) = 9.045 hours

So all the coordinates are:
1: (10, 11.11)
2: (8.852, 10.08)
3: (7.705, 9.045)

Now these answers, to me, make sense if your compare them to the minkowski space-time diagrams for this situation. THIS is my question for you guys: when using the above inverse lorentz transformations that i have used, I thought the 'v' represented the velocity of the SSBS frame w.r.t the Swarm frame, i.e. v = -0.9c (since if we consider the swarm stationary, then the Earth is moving towards them at a speed of 0.9c), However when i plug this value for v into the transformation equations, i get nonsensical answers (at least to me). So am i using the transformations correctly, with v = 0.9c? If im not, why do the answers look 'unintuitive and if i am using them correctly, why is v=0.9c not v=-0.9c? (could it be that the form of inverse transforms that i am using have already taken into account that the sign of the velocity would be reverse?)

Thank you for your time and attention!

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TSny
Homework Helper
Gold Member
So first I calculated the contracted distance between the first beacon and Earth according to the swarm of meteors, and this came out to be 4.3589 light hours. From this i calculated the proper time (the time in the swarm ref frame) between the two events (passing the first beacon and arriving at Earth) this came out to be 4.843 hours. Then I found the time dilation in the SSBS frame for the two events, and it was 11.11 hours.

Next thing i did was find the co ordinates of all the meteors in the swarm frame when the first meteor arrived at Earth. These were their co ordinates in the form (x,t) in light hours and hours;
Meteor 1: (0, 4.843)
meteor 2: (-0.5, 4.843)
meteor 3: (-1, 4.843)
Since the spatial origin is always at the first meteor in the swarm frame.
OK.

Now all i need to do is use the inverse lorentz transformations, right?
In question (d) it appears to me that you are looking for the location of the three meteors at the same instant of time in the SSBS frame; i.e., simultaneously in the SSBS frame at the instant the first meteor reaches earth.

The three events you listed above are simultaneous in the meteor frame. Are they simultaneous events in the SSBS frame?

Sorry question (c) asks for the spacetime coordinates of all the meteors at the instant the first meteor arrives at Earth in the swarm frame. And question (d) asks to convert the co ordinates in (c) to the SSBS frame.

Im fairly sure that my answer to (d) is correct im just not sure why it is because I have used +0.9c instead of -0.9c. My actual question in this post is stated in the last paragraph (and I apologize for the extraneous length!).

TSny
Homework Helper
Gold Member
In the standard way of writing the Lorentz transformation equations, v represents the magnitude (absolute value) of the relative velocity of the two frames. So, v is a positive number in all of the equations. But, the equations contain different signs in front of v, depending on whether you are transforming from the unprimed frame to the primed frame, or from the primed to the unprimed frame. This difference in sign takes care of the fact that the primed frame moves in the positive x direction relative to the unprimed frame while the unprimed frame moves in the negative x' direction relative to the primed frame.

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Aahhh I had a feeling that was the case. Thank you thats exactly what I needed to hear and exactly what our lecturer failed to tell us.

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